Total moment of inertia of a two particle system


by Linus Pauling
Tags: inertia, moment, particle
Linus Pauling
Linus Pauling is offline
#1
Oct24-09, 04:49 PM
P: 190
1. Find the moment of inertia I_x of particle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia I_y of particle a with respect to the y axis, and the moment of inertia I_z of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes).

Particle a is located 3r from the y axis, and particle b is located r away.




2. I = mr^2
I = SUMM(m_i*r_i^2)




3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Doc Al
Doc Al is offline
#2
Oct24-09, 04:54 PM
Mentor
Doc Al's Avatar
P: 40,889
Quote Quote by Linus Pauling View Post
Particle a is located 3r from the y axis, and particle b is located r away.
Give the complete coordinates and the mass of each particle.
Linus Pauling
Linus Pauling is offline
#3
Oct24-09, 04:57 PM
P: 190
Particle a:

distance from y axis = 3r
distance from x axis = r

Particle b:

distance from y axis = r
distance from x axis = -4r

It doesn't say anything explicitly about mass, and so I think we assume each particle has mass m.

Doc Al
Doc Al is offline
#4
Oct24-09, 05:02 PM
Mentor
Doc Al's Avatar
P: 40,889

Total moment of inertia of a two particle system


Quote Quote by Linus Pauling View Post
3. I_a = 9mr^2
I_b = mr^2

Total I = 10mr^2

No idea what I am doing wrong
Looks like you've found I_y. What about I_x?
Linus Pauling
Linus Pauling is offline
#5
Oct24-09, 05:06 PM
P: 190
I_x would be m(r)^2 + m(-4r)^2 = mr^2 + 16mr^2 = 17mr^2.

Am I simply adding that to I_y to get I_total? It asks for total I with respect to the y axis, though, why would I need I_x?
Linus Pauling
Linus Pauling is offline
#6
Oct24-09, 05:10 PM
P: 190
Shoot, just realized I posted to wrong question. Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Doc Al
Doc Al is offline
#7
Oct24-09, 05:25 PM
Mentor
Doc Al's Avatar
P: 40,889
Quote Quote by Linus Pauling View Post
Here's what I am trying to answer:

Find the total moment of inertia I of the system of two particles shown in the diagram with respect to the y axis
Assuming the diagram matches your description, your answer seems correct to me. (Post the diagram, if you can.)
Linus Pauling
Linus Pauling is offline
#8
Oct24-09, 05:26 PM
P: 190
Doc Al
Doc Al is offline
#9
Oct24-09, 05:34 PM
Mentor
Doc Al's Avatar
P: 40,889
Looks good to me. I say your answer is correct.
Linus Pauling
Linus Pauling is offline
#10
Oct24-09, 05:35 PM
P: 190
Ok, I requested the correct answer: 11mr^2

WTF?!
Doc Al
Doc Al is offline
#11
Oct24-09, 05:37 PM
Mentor
Doc Al's Avatar
P: 40,889
Quote Quote by Linus Pauling View Post
Ok, I requested the correct answer: 11mr^2
Sounds bogus to me. You might want to mention this to your instructor.
Linus Pauling
Linus Pauling is offline
#12
Oct24-09, 05:39 PM
P: 190
Ok, now:

Using the formula for kinetic energy of a moving particle K=\frac{_1}{^2}mv^2, find the kinetic energy K_a of particle a and the kinetic energy K_b of particle b.
Express your answers in terms of m, omega, and r separated by a comma.

I know I need to calculate linear speed, how do I do that? I know it's v=omega*r but I don't see how to put this together.....
Doc Al
Doc Al is offline
#13
Oct24-09, 05:42 PM
Mentor
Doc Al's Avatar
P: 40,889
Quote Quote by Linus Pauling View Post
I know I need to calculate linear speed, how do I do that?
What's the relationship between tangential speed and angular speed for something going in a circle?
Linus Pauling
Linus Pauling is offline
#14
Oct24-09, 05:44 PM
P: 190
Just edited before you replied: v=omega*r

But what are omega and r here?!
Linus Pauling
Linus Pauling is offline
#15
Oct24-09, 05:49 PM
P: 190
Particle A:

Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka.............
Linus Pauling
Linus Pauling is offline
#16
Oct24-09, 05:58 PM
P: 190
Ok I got Kb = m(omega*r)^2

What is total kinteic eneregy?
Doc Al
Doc Al is offline
#17
Oct25-09, 09:02 AM
Mentor
Doc Al's Avatar
P: 40,889
Quote Quote by Linus Pauling View Post
Ka = 0.5m(3*omega*r)^2

Kb = 0.5m(omega*r)^2

Apparently Kb is wrong but I did it the same way as Ka.............
These both look OK to me.

Quote Quote by Linus Pauling View Post
Ok I got Kb = m(omega*r)^2
How did you get that? (Are you posting this problem exactly as given, word for word?)

What is total kinetic energy?
Just add up the kinetic energy of each mass.


Register to reply

Related Discussions
moment of inertia for a system General Engineering 2
I need to to find the moment of inertia for the system Introductory Physics Homework 1
Moment of Inertia of a System Calculus & Beyond Homework 6
Pulley system with Moment of Inertia Introductory Physics Homework 0
Calculating moment of inertia of a system of balls Introductory Physics Homework 4